# Thread: expansion, binomial

1. ## expansion, binomial

In the binomial expansion of $(2k+x)^n$ where k is a constant and n is a positive integer, the coefficient of $x^2$ is equal to the coefficient of $x^3$ .

(a) Prove that n=6k+2.
$(2k+x)^n = 2k^n + ^n\mathrm{C}_1 (2k)^{n-1}x +^n\mathrm{C}_2 (2k)^{n-2} x^2 +^n\mathrm{C}_3 (2k)^{n-3}x^3 +....$

Coefficient of $x^2$ = coefficient of $x^3$

$^n\mathrm{C}_2(2k)^{n-2} = ^n\mathrm{C}_3(2k)^{n-3}$

$\frac{n!}{ (n-2)!2!} (2k)^{n-2} = \frac{n!}{ (n-3)!3!} (2k)^{n-3}$

I dont know how to simplify this further to get $n=6k+2$

Any help appreciated,

Thanks.

2. Originally Posted by Tweety
I dont know how to simplify this further to get $6k=n-2$
$\begin{gathered}
6k = n - 2 \hfill \\
6k + 2 = n - 2 + 2 \hfill \\
6k + 2 = n \hfill \\
\end{gathered}$

3. Originally Posted by Tweety
$(2k+x)^n = 2k^n + ^n\mathrm{C}_1 (2k)^{n-1}x +^n\mathrm{C}_2 (2k)^{n-2} x^2 +^n\mathrm{C}_3 (2k)^{n-3}x^3 +....$

Coefficient of $x^2$ = coefficient of $x^3$

$^n\mathrm{C}_2(2k)^{n-2} = ^n\mathrm{C}_3(2k)^{n-3}$

$\frac{n!}{ (n-2)!2!} (2k)^{n-2} = \frac{n!}{ (n-3)!3!} (2k)^{n-3}$

I dont know how to simplify this further to get $6k=n-2$

Any help appreciated,

Thanks.

$\frac{n!}{ (n-2)!2!} (2k)^{n-2} = \frac{n!}{ (n-3)!3!} (2k)^{n-3}$

$\frac{n!}{ (n-2)!2!} (2k)^{n-2} - \frac{n!}{ (n-3)!3!} (2k)^{n-3} =0$

Now lets factor

$\frac{n!}{ (n-2)!2!} (2k)^{n-2} - \frac{n!}{ (n-3)!3!} (2k)^{n-3} =0 \iff$

$n!(2k)^{n-3}\left( \frac{2k}{(n-2)!2!}-\frac{1}{(n-3)!3!}\right)=0$

Now by the zero factor rule we get

$\left( \frac{2k}{(n-2)!2!}-\frac{1}{(n-3)!3!}\right)=0 \iff \left( \frac{2k}{(n-2)!2!}=\frac{1}{(n-3)!3!}\right)$

$2k=\frac{(n-2)!2!}{(n-3)!3!} \iff 2k=\frac{(n-2)(n-3)!2!}{(n-3)!3\cdot 2!}$

$2k=\frac{(n-2)}{3} \iff 6k=n-2$

4. ^^thanks, Althought I dont know what the 'zero factor rule' is? So I am not really following how you got the desired expression?

5. Originally Posted by Tweety
^^thanks, Althought I dont know what the 'zero factor rule' is? So I am not really following how you got the desired expression?
You probably know it by a different name

It states that if
$a\cdot b=0$ then $a=0 \mbox{ or } b=0$

It is what you use to solve equations by factoring and setting each factor equal to zero.

We know that exponentials and factorials are never zero so the other factor must be zero.

6. Originally Posted by TheEmptySet
You probably know it by a different name

It states that if
$a\cdot b=0$ then $a=0 \mbox{ or } b=0$

It is what you use to solve equations by factoring and setting each factor equal to zero.

We know that exponentials and factorials are never zero so the other factor must be zero.
Oh yes I do know that, thanks. Also how comes you are able to factor out $(2k)^{n-3}$ ? Becasue if you multiply it back you wont end up getting $(2k)^{n-2}$
Can you explain ? Thanks!

7. Originally Posted by Tweety
Oh yes I do know that, thanks. Also how comes you are able to factor out $(2k)^{n-3}$ ? Becasue if you multiply it back you wont end up getting $(2k)^{n-2}$
Can you explain ? Thanks!
$(2k)^{n-3}(2k)=(2k)^{n-3+1}=(2k)^{n-2}$

Remember when we multiply things with the same base we add the exponents